For the following chemical reaction, determine the amount of water in grams produced when 2.05 x 10 molecules of H react with 1.33 x 10 molecules of O to produce water 2H(g) + O(g) → 2HO(l)

6.13 grams of H2O is produced

Explanation:

For the following chemical reaction, determine the amount of water in grams produced when 2.05 x 10^23 molecules of H2 react with 1.33 x 10^23 molecules of O2 to produce water

Step 1: Data given

Number of H2 molecules = 2.05 *10^23 molecules

Number of O2 molecules = 1.33 * 10^23 molecules

Number of Avogadro = 6.022*10^23 / mol

Molar mass H2 = 2.02 g/mol

Molar mass O2 = 32.0 g/mol

Molar mass H2O = 18.02 g/mol

Step 2: The balanced equation

2H2(g) + O2(g) → 2H2O(l)

Step 3: Calculate moles

Moles H2 = 2.05 * 10^23 / 6.022 * 10^23

Moles H2 = 0.340 moles

Step 4: Calculate moles O2

Moles O2 = 1.33 * 10^23 molecules / 6.022*10^23

Moles O2 = 0.221 moles

Step 5: Calculate limiting reactant

For 2 moles H2 we need 1 mol O2 to produce 2 moles H2O

H2 is the limiting reactant. It will completely be consumed (0.340 moles).

O2 is in excess. There will react 0.340/2 = 0.170 moles

There will remain 0.221 - 0.170 = 0.051 moles

Step 6: Calculate moles H2O

For 2 moles H2 we need 1 mol O2 to produce 2 moles H2O

For 0.340 moles we'll have 0.340 moles H2O

Step 7: Calculate mass H2O

Mass H2O = moles H2O * molar mass H2O

Mass H2O = 0.340 moles * 18.02 g/mol

Mass H2O = 6.13 grams

6.13 grams of H2O is produced

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.

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